Question

A pizza shop owner wishes to find the 95% confidence interval of the true mean cost of a large plain pizza. How large should the sample be if she wishes to be accurate to within $0.17? A previous study showed that the standard deviation of the price was $0.42. Round your final answer up to the next whole number

Answer 1

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Solution

standard deviation =s =   =0.42

Margin of error = E = 0.17

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96 ( Using z table ( see the 0.025 value in standard normal (z) table corresponding z value is 1.96 )

sample size = n = [Z/2* / E] 2

n = ( 1.96* 0.42/ 0.17 )2

n =23.448

Sample size = n =24